Symmetricity of the dirac delta function

Question

The common property of the delta function is:

$\int_{-\infty}^{\infty}f(x)\delta(x-a)dx = f(a)$

However the proof of the Greens Theorem states that

$\int_{-\infty}^{\infty}f(a)\delta(x-a)da = f(x)$

How are these two equivalent? In the second equation should we not switch $x$ and $a$?

Answer 1

Denote $a$ by $y$ in the second equation. Then we have

$$\int_{-\infty}^{\infty}f(y)\delta(x-y)\ dy=\int_{-\infty}^{\infty}f(y)\delta(-(y-x))\ dy.$$

Since $\delta(x)$ is an even "function" we have

$$\delta(-(y-x))=\delta(y-x).$$

So, $$\int_{-\infty}^{\infty}f(y)\delta(y-x)\ dy=f(x).$$

Now, change back $y$ to $a$ and play it again.

Answer 2

If you exchange the variables $x$ and $a$ then you get

$$ \int_{-\infty}^{\infty}{f(x) \delta(a-x)dx} = f(a), $$

but if you also use that the delta Dirac function is even, it is equivalent to your first equation.